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Backtracking-3 Done #1075

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74 changes: 74 additions & 0 deletions n-queens.py
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Original file line number Diff line number Diff line change
@@ -0,0 +1,74 @@
''' Time Complexity : O(n * n!) ;
Space Complexity : O(n^2 + n)
Did this code successfully run on Leetcode : Yes
Any problem you faced while coding this : No


Approach - Place the queen if safe , go to next row. if cannot place in next row, backtrack by making 1->0
'''

class Solution:
def solveNQueens(self, n: int) -> List[List[str]]:
self.mat=[[False for _ in range(n)] for _ in range(n)]
self.result = []

def helper(i, j, n):
#base
if i == n:
li = []
for r in range(n):
s = ""
for c in range(n):
if self.mat[r][c]:
s += "Q"
else:
s += "."
li.append(s)
self.result.append(li)
return

for j in range(n):
if self.isSafe(i,j,n):
self.mat[i][j] = True
helper(i+1,j,n)
self.mat[i][j] = False

helper(0,0,n)
return self.result


def isSafe(self,i, j,n):
r,c = i, j
#up check
while r>=0:
if self.mat[r][c]:
return False
r -= 1

r,c = i, j
#right check diagonal
while r>=0 and c<n:
if self.mat[r][c]:
return False
r -= 1
c += 1

r,c = i, j
#right left diagonal
while r>=0 and c>=0:
if self.mat[r][c]:
return False
r -= 1
c -= 1
return True











41 changes: 41 additions & 0 deletions word-search.py
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''' Time Complexity : O(m * n * 4^L) ; L - len of word
Space Complexity : O(L)
Did this code successfully run on Leetcode : Yes
Any problem you faced while coding this : No

Approach : First do left pass to calculate running product.
Then second iteration from right to left and updating the new product
'''

class Solution:
def exist(self, board: List[List[str]], word: str) -> bool:
rows, cols = len(board),len(board[0])
dirs = [(-1,0),(0,1),(1,0),(0,-1)]

if rows==1 and cols == 1 and board[0][0] == word:
return True

def dfs(board, word, i, j, s):
#base
if s == len(word):
return True
if board[i][j] != word[s]:
return False
#logic
#action
char = board[i][j]
board[i][j] = "#"
for dir in dirs:
r = dir[0] + i
c = dir[1] + j
if 0<=r<rows and 0<=c<cols:
if dfs(board, word, r, c, s+1):
return True
#backtrack
board[i][j] = char

for i in range(rows):
for j in range(cols):
if dfs(board, word, i, j, 0):
return True
return False